Solve {l}{3x+2y+2z=3}{x+2y-z=5}{2x-4y+z=0}

$$\left. \begin{array} { l } { 3 x + 2 y + 2 z = 3 } \\ { x + 2 y - z = 5 } \\ { 2 x - 4 y + z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=\frac{1}{2}=0.5$$

$$z=-2$$

Short Steps Using Substitution

Reorder the equations.

$$x+2y-z=5 3x+2y+2z=3 2x-4y+z=0$$

Solve x+2y-z=5 for x.

$$x=-2y+z+5$$

Substitute -2y+z+5 for x in the second and third equation.

$$3\left(-2y+z+5\right)+2y+2z=3 2\left(-2y+z+5\right)-4y+z=0$$

Solve these equations for y and z respectively.

$$y=\frac{5}{4}z+3 z=-\frac{10}{3}+\frac{8}{3}y$$

Substitute \frac{5}{4}z+3 for y in the equation z=-\frac{10}{3}+\frac{8}{3}y.

$$z=-\frac{10}{3}+\frac{8}{3}\left(\frac{5}{4}z+3\right)$$

Solve z=-\frac{10}{3}+\frac{8}{3}\left(\frac{5}{4}z+3\right) for z.

$$z=-2$$

Substitute -2 for z in the equation y=\frac{5}{4}z+3.

$$y=\frac{5}{4}\left(-2\right)+3$$

Calculate y from y=\frac{5}{4}\left(-2\right)+3.

$$y=\frac{1}{2}$$

Substitute \frac{1}{2} for y and -2 for z in the equation x=-2y+z+5.

$$x=-2\times \frac{1}{2}-2+5$$

Calculate x from x=-2\times \frac{1}{2}-2+5.

$$x=2$$

The system is now solved.

$$x=2 y=\frac{1}{2} z=-2$$